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Today Pari and Arya are playing a game called Remainders.
Pari chooses two positive integer x and k, and tells Arya k but not x. Arya have to find the value . There are n ancient numbersc1, c2, ..., cn and Pari has to tell Arya
if Arya wants. Given k and the ancient values, tell us if Arya has a winning strategy independent of value of x or not. Formally, is it true that Arya can understand the value
for any positive integer x?
Note, that means the remainder of x after dividing it by y.
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000 000) — the number of ancient integers and value k that is chosen by Pari.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 1 000 000).
Output
Print "Yes" (without quotes) if Arya has a winning strategy independent of value of x, or "No" (without quotes) otherwise.
Sample Input
4 5 2 3 5 12
Yes
2 7 2 3
No
Hint
In the first sample, Arya can understand because 5 is one of the ancient numbers.
In the second sample, Arya can't be sure what is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7.
思路：首先，根据剩余定理，如果我们想知道x%m等于多少，当且仅当我们知道x%m1,x%m2..x%mr分别等于多少，其中m1*m2...*mr=m，并且mi相互互质，即构成独立剩余系。令m的素数分解为m=p1^k1*p2^k2...*pr^kr，如果任意i，都有pi^ki的倍数出现在集合中，那么m就能被猜出来。
这个问题等价于问LCM(ci)%m是否等于0
所以只要求出LCM(ci)即可，不过要边求lcm，边和m取gcd，防止爆int
ac代码：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
const int maxn = 1e5+100;
typedef long long LL;
LL gcd(LL a,LL b) { return b==0 ? a:gcd(b,a%b);}
LL lcm(LL a,LL b){ return a/gcd(a,b)*b; }
int main()
{
int n,k,a,i;
scanf("%d%d",&n,&k);
LL ans=1;
for(i=1;i<=n;i++)
{
scanf("%d",&a);
ans=lcm(ans,a)%k;
}
printf("%s\n",ans==0 ? "Yes":"No");
return 0;
}
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